The projection of the line x+1-1=y2=z-13 on the plane x-2y+z=6 is the line of intersection of this plane with the plane

Given line x+1-1=y2=z-13 And plane is x-2y+z=6 Equation of the plane through (-1,0,1) is a(x+1)+b(y-0)+c(z-1)=0…….(1) Which is parallel to the given line and perpendicular to the given plane ⇒-a+2 b+3 c=0 and a-2b+c=0⇒c=0,a=2b∴ From (1) equation of the required plane is 2x+y+2=0